A174711 Composites of the form 2*n^n + 1 = A216147(n).

9, 55, 513, 6251, 93313, 1647087, 33554433, 774840979, 20000000001, 570623341223, 605750213184507, 22224013651116033, 875787780761718751, 36893488147419103233, 1654480523772673528355, 3956839311320627178247959

Offset: 1

Author

Michel Lagneau, Mar 27 2010

Keywords

• nonn

Comments

If p = n+2 is prime, then p divides 2*n^n + 1. Proof: Let p = n+2 prime. Then, according to Fermat's theorem, n^(p-1) == 1 (mod p). Because p-1 = n+1, n^(n+1) == 1 (mod p), and with n = p-2 == -2 (mod p), we obtain successively: n*n^n == 1 (mod p), -2*n^n == 1 (mod p), 2*n^n == -1 (mod p) => p divides 2*n^n + 1.

Examples

			a(2) = 9 = 3^2, a(3) = 55 = 5*11, a(4) = 513 = 3 ^ 3 * 19.
		

References

• T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
• J. M. De Koninck, A. Mercier, 1001 problemes en theorie classique des nombres, Ellipses 2004, p. 52.
• G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 2.

Links

• C. K. Caldwell, Composite Numbers

Crossrefs

Complement of A216148 in A216147. - M. F. Hasler, Sep 02 2012

Programs

Maple

with(numtheory):for n from 0 to 50 do: x:=2*n^n + 1 : if type(x,prime)=false then print (x):else fi:od:

Mathematica

Select[Table[2n^n+1,{n,20}],CompositeQ] (* Harvey P. Dale, Jun 21 2015 *)